4 Sobolev spaces, trace theorem and normal derivative

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1 4 Sobolev spaces, trace theorem and normal derivative Throughout, n will be a sufficiently smooth, bounded domain. We use the standard Sobolev spaces H 0 ( n ) := L 2 ( n ), H 0 () := L 2 (), H k ( n ), H k () (k positive integer). Note that all these spaces are based on the use of weak derivatives up to order k. We will use the Fourier transform to redefine the norms in these spaces. Recall that the Fourier transform F is defined by (there are different normalisations possible) ˆv(ξ) := Fv(ξ) := e i2πξ x v(x)dx (ξ n ). n Since ˆv(ξ) = e i2πξ x v(x)dx e i2πξ x v(x) dx = v(x) dx n n n it follows that ˆv is well-defined whenever v L 1 ( n ). The inversion formula for the Fourier transform is F 1ˆv(x) := e i2πξ xˆv(ξ)dξ. n One finds the following properties: If v, ˆv L 1 ( n ) then F 1 Fv = v = FF 1 v wherever v is continuous. F generalises to a bounded linear mapping and there holds F : L 2 ( n ) L 2 ( n ) (Fϕ, Fv) = (ϕ,v) = (F 1 ϕ, F 1 v) ϕ,v L 2 ( n ), i.e., F is a unitary isomorphism. This property is known as Plancherel s theorem. The symbol (, ) denotes the L 2 inner product on n and will be used throughout, also for its extension by duality. When referring to the inner product on a subset of n, e.g. on, we add this subset as an index, e.g. (, ). A conclusion from Plancherel s theorem is the relation v L2 ( n ) = ˆv L2 ( n ) v L 2 ( n ). Example 4.1 Consider the one-dimensional case, i.e., n = 1. There holds v L2 () = F(v ) L2 (), and for any v H 1 () with compact support we obtain F(v )(ξ) = e i2πxξ v (x)dx = v(x)e i2πxξ i2πξ e i2πxξ v(x)dx = i2πξ ˆv(ξ). x= 18

2 Therefore, and so that v L2 () = i2πξ ˆv(ξ) L2 () = 2π ξ ˆv(ξ) L2 () v 2 H 1 () = v 2 L 2 () + v 2 L 2 () = ˆv 2 L 2 () + 4π2 ξ ˆv 2 L 2 () = v H 1 () are equivalent norms. and ( (1 + 4π 2 ξ 2 ) ˆv(ξ) 2 dξ, (1 + ξ 2 ) ˆv(ξ) 2 dξ) 1/2 = (1 + ξ 2 ) 1/2ˆv L2 () This example easily generalises to higher dimensions (n > 1). Moreover, it leads us to the definition of Sobolev spaces on n for any positive real order. Definition 4.1 For s > 0 we define { H s ( n ) := v L 2 ( n ); (1 + ξ 2 ) s/2ˆv } L2 ( n ) < with norm v H s ( n ) := (1 + ξ 2 ) s/2ˆv L2 ( n ). As in Example 4.1 one sees that, for integer s, this norm is equivalent to the usual one (based on derivatives). For non-integer s, H s ( n ) is called a fractional order Sobolev space. We are now in a position to analyse the trace operator in the half-space case. Consider the situation given in Figure 4.1. For x = (x 1,...,x n ) n we denote x := (x 1,...,x n 1 ). Then we define for v C 0 (n ) its trace onto the hyperplane n 1 {0} by γ 0 v(x ) := v(x,x n = 0), x n 1. Theorem 4.1 (trace theorem, half-space case) For s > 1/2 there exists a unique extension of γ 0 to a bounded linear operator γ 0 : H s ( n ) H s 1/2 ( n 1 ). Proof. By density it suffices to consider v C0 (n ). By the Fourier inversion formula we find that γ 0 v(x ) = e i2πx ξˆv(ξ)dξ ( = e i2πx ξ ˆv(ξ)dξ = ˆv(ξ,ξ n )dξ n )e i2πx ξ dξ. n xn=0 n n 1 Therefore, F(γ 0 v)(ξ ) = ˆv(ξ,ξ n )dξ n = (1 + ξ 2 ) s/2 (1 + ξ 2 ) s/2ˆv(ξ,ξ n )dξ n 19

3 x n x n=0 x =(x 1,...,x n 1 ) Figure 4.1: The trace in the half-space case. and an application of the Cauchy-Schwarz inequality yields F(γ 0 v)(ξ ) 2 (1 + ξ 2 ) s dξ n (1 + ξ 2 ) 2 ˆv(ξ,ξ n ) 2 dξ n. Now, by the substitution ξ n = (1 + ξ 2 ) 1/2 t, M s (ξ ) := (1 + ξ 2 ) s dξ n dξ n = (1 + ξ 2 + ξ n 2 ) s 1 dt = (1 + ξ 2 ) s 1/2 (1 + t 2 < iff s > 1/2. ) s Therefore, we can bound (1 + ξ 2 ) s 1/2 F(γ 0 v)(ξ ) 2 C s (1 + ξ 2 ) 2 ˆv(ξ) 2 dξ n for a constant C s depending on s, and integration with respect to ξ yields γ 0 v H s 1/2 ( n 1 ) C1/2 s v H s ( n ). So far we have dealt with Sobolev spaces on n. For boundary value problems on Lipschitz domains this is obviously not enough. Definition 4.2 Let n be a Lipschitz domain. For s 0 we introduce the following spaces: H s () := H s ( n ) with norm v H s () := inf V H V =v s ( n ), H s 0() := C 0 () H s () with norm H s (), 20

4 and H s () := {v H s (); v 0 H s ( n )} with norm v Hs () := v0 H s ( n ) where v 0 denotes the extension of v by 0 onto n \. For s < 0 we define H s () := ( H s ()) (dual space) with operator norm and H s () := ( H s ()) (dual space) with operator norm. Remark 4.1 One can show that, for s > 0, Hs () = H0 s () if s integer + 1/2. In the cases s = integer + 1/2 the spaces are different, Hs () H0 s () in general. Without going into the details, we mention that on a Lipschitz surface or boundary Γ all the above spaces can be defined analogously when s 1. To this end one uses a partition of unity and local transformations onto subsets of n 1. Higher order spaces require more regularity of Γ. The trace theorem can be generalised to Lipschitz domains. Theorem 4.2 (trace theorem, general form) Let n be a bounded Lipschitz domain with boundary Γ. (i) For 1/2 < s < 3/2, γ 0 has a unique extension to a bounded linear operator γ 0 : H s () H s 1/2 (Γ). (ii) For any s (1/2,3/2) and any v H s 1/2 (Γ) there exists V := Ev H s () such that γ 0 (V ) = v and Ev H s () C s () v H s 1/2 (Γ) v H s 1/2 (Γ). Remark 4.2 Part (ii) of Theorem 4.2 means that γ 0 has a right-inverse: which is continuous, and that ) is surjective, i.e., γ 0 (H s () operator. v = γ 0 V = γ 0 Ev γ 0 : H s () H s 1/2 (Γ) = H s 1/2 (Γ). Of course, this right-inverse E is an extension Having the trace operator at hand we can now make an interpretation of the Dirichlet boundary condition. Studying the Poisson equation with Dirichlet boundary condition u = f in, u Γ = g 21

5 we conclude two things. First, the equation u Γ = g means that γ 0 u = g in H 1/2 (Γ) since the variational formulation of the Poisson equation is posed in H 1 () (subject to the Dirichlet condition). Second, the Dirichlet condition makes sense only for g H 1/2 (Γ). If g H 1/2 (Γ) then there does not exist a solution u H 1 () of the given boundary value problem. This is a conclusion of the surjectivity of the trace operator. Besides the trace operator γ 0, in 1 we were concerned about the definition of the normal derivative n v of a function v H 1 (). We now deal with this operator. The origin for the definition of the normal derivative is the first Green s formula, in the form v w = v w n v w. This leads us to the definition of n v for v H 1 () by n v,w Γ := v W + Γ v W where W H 1 () is any extension of w H 1/2 (Γ). The notation Φ,ϕ Γ means the application of the functional Φ to ϕ defined on Γ, in this case it is the duality between H 1/2 (Γ) and H 1/2 (Γ). For Φ, ϕ L 2 (Γ) it is simply the L 2 (Γ)-inner product between Φ and ϕ. Lemma 4.1 n : {v H 1 (); v H 1 ()} H 1/2 (Γ) is well-defined and continuous when defining v W := ( v,w) as duality between H 1 () and H 1 (). Proof. (i) First we show that the definition of n v,w Γ is independent of the extension W of w. Let W 1, W 2 H 1 () be two extensions of w, i.e., γ 0 W 1 = γ 0 W 2 = w. Then v (W 1 W 2 ) + v (W 1 W 2 ) = 0 by the second Green identity since W 1 W 2 H 1 0 (). This proves that nv,γ 0 (W 1 W 2 ) Γ = 0 as wanted. (ii) Now we show the boundedness of n. Let E : H 1/2 (Γ) H 1 () denote the extension 22

6 operator from Theorem 4.2(ii). We estimate n v H 1/2 (Γ) = = C sup w H 1/2 (Γ)\{0} sup w H 1/2 (Γ)\{0} n v,w Γ w H 1/2 (Γ) C v Ew + v Ew C Ew H 1 () sup w H 1/2 (Γ)\{0} sup W H 1 ()\{0} n v,w Γ Ew H 1 () v H C sup 1 () W H 1 () + v H 1 () W H 1 () W H 1 ()\{0} W H 1 () = C ( v H 1 () + v H 1 ()). v W + v W W H 1 () Remark 4.3 v L 2 () implies v H 1 (). 23